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Find the sum of all two digit numbers which when divided by $4$ yields $1$ as remainder.

$\begin{array}{1 1}1196 \\ 1210 \\ 1312 \\ 1218 \end{array} $

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  • $n^{th}$ term of an A.P.=$a+(n-1)d$
  • Sum of $n$ terms of an A.P=$\large\frac{n}{2}$$(t_n+a)$
Two digit numbers that yields remainder $1$ when divided by $4$ forms sequence
This sequence is an A.P with first term $a=13$ common difference $d=4$ and
or $n=22$
$\therefore$ Sum of the required numbers =$13+17+21+.......97$
$=\large\frac{n}{2}$$(t_n+a)=11(97+13)=11\times 110=1210$
answered Mar 29, 2014 by rvidyagovindarajan_1

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